Solution:

Let $EF$ denote the number of students speaking English and French. Similarly define $ES$, $FS$, $E$, $F$, $S$, $EFS$. Then $ES + EF + E + EFS = 50$, $EF + FS + F + EFS = 50$. Subtracting: $ES - F = FS - E$. Similarly, $ES - F = EF - S$.

Pair off members of $FS$ with members of $E$. Similarly, members of $ES$ with $F$, and members of $EF$ with $S$. The resulting pairs have one person speaking each language. If $\{ES\} = \{F\}$, then the only remaining students are those in $EFS$, who speak all three languages. We thus have a collection of units (pairs or individuals) each containing one speaker of each language.

If $\{ES\} < \{F\}$, then after the pairing off we are left with equal numbers of members of $E$, $F$, and $S$. These may be formed into triplets, with each triplet containing one speaker of each language. As before we also have the students in $EFS$. Again, we have partitioned the student body into units with each unit containing one speaker of each language.

If $\{ES\} > \{F\}$, then after the pairing off, we are left with an equal number of members of $ES$, $FS$ and $EF$. These may be formed into triplets, with each triplet containing two speakers of each language. So, in this case we partition the student body into units with each unit containing either one speaker of each language, or two speakers of each language.

Finally, we may divide the units into $5$ groups with $10$ speakers of each language in each group.